What is the integral of t(1-t^2)^{10} ?

The integral of t(1-t^2)^{10} is 1/22 (-1+t^2)^{11}+C

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integral of t(1-t^2)^{10}

\int t (1 - t^{2})^{10} d t

\frac{1}{22} (- 1 + t^{2})^{11} + C

Solution steps

\int t (1 - t^{2})^{10} d t

Factor

1 - t^{2} : - (- 1 + t^{2})

= \int t (- (- 1 + t^{2}))^{10} d t

(- (- 1 + t^{2}))^{10} = (- 1 + t^{2})^{10}

= \int t (- 1 + t^{2})^{10} d t

Apply u-substitution

= \int \frac{u ^{10}}{2} d u

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= \frac{1}{2} \cdot \int u^{10} d u

Apply the Power Rule

= \frac{1}{2} \cdot \frac{u ^{11}}{11}

Substitute back

u = - 1 + t^{2}

= \frac{1}{2} \cdot \frac{( - 1 + t ^{2} ) ^{11}}{11}

Simplify

\frac{1}{2} \cdot \frac{( - 1 + t ^{2} ) ^{11}}{11} : \frac{1}{22} (- 1 + t^{2})^{11}

= \frac{1}{22} (- 1 + t^{2})^{11}

Add a constant to the solution

= \frac{1}{22} (- 1 + t^{2})^{11} + C

What is the integral of t(1-t^2)^{10} ?
The integral of t(1-t^2)^{10} is 1/22 (-1+t^2)^{11}+C